Question

Someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi. We think the actual amount is lower than that and want to run the test at an alpha level of 5%. What would our sample size need to be if we want to reject the null hypothesis if the sample mean is at or below 1,997.2956?

Answer 1

The sample size must be greater than 37 if we want to reject the null hypothesis.

Explanation:

We are given that someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi.

Also, we are given a level of significance of 5%.

Let
`\mu` = mean breaking strength of their climbing rope

SO, Null Hypothesis,
`H_0` :
`\mu` = 2,000 psi {means that the mean breaking strength of their climbing rope is 2,000 psi}

Alternate Hypothesis,
`H_A` :
`\mu` < 2,000 psi {means that the mean breaking strength of their climbing rope is lower than 2,000 psi}

Now, the test statistics that we will use here is One-sample z-test statistics as we know about population standard deviation;

T.S. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = ample mean strength = 1,997.2956 psi

`\sigma` = population standard devaition = 10 psi

n = sample size

Now, at the 5% level of significance, the z table gives a critical value of -1.645 for the left-tailed test.

So, to reject our null hypothesis our test statistics must be less than -1.645 as only then we have sufficient evidence to reject our null hypothesis.

SO, T.S. < -1.645 {then reject null hypothesis}

`(\bar X-\mu)/((\sigma)/(√(n) ) ) < -1.645`

`(1,997.2956-2,000)/((10)/(√(n) ) ) < -1.645`

`((1,997.2956-2,000)/(10)) * {√(n) } } < -1.645`

`-0.27044 * √(n)< -1.645`

`√(n)> (-1.645)/(-0.27044)`

`√(n)>6.083`

n > 36.99 ≈ 37.

SO, the sample size must be greater than 37 if we want to reject the null hypothesis.